Consider a random variable $V$ with distribution function $F$ and density function $f$ with support $[\underline{v},\overline{v}]$, where $0\leq\underline{v}<\overline{v}$. The mean is $\mu$. Here $f$ is assumed to be differentiable on its support and log-concave (i.e. $\ln \circ f$ is a concave function).
I would like to prove (or disprove) that
$$\tag{1}\phi(\mu)=\mu F(\mu)^2-\int_{\mu}^{\overline{v}}F(v)[1-F(v)]dv\geq 0$$
for all such probability distributions.
I have not been able to find any examples of distributions for which inequality $(1)$ is violated (I have tried the beta and Kumaraswamy distributions on $[0,1]$ with many different combinations of parameters).
An example of an $f$ that is not log concave for which $\phi(\mu)<0$ would also be helpful (as I am not entirely sure whether the log-concavity is relevant).
Inequality $(1)$ can also be written as (using integration by parts):
$$\begin{align}&E[V\mid V\geq \mu]\cdot P(V\geq \mu)+\mu\cdot P[V\leq \mu]-E[\hat{V}\mid \hat{V}\geq \mu]\cdot P(\hat{V}\geq \mu)\\ &=\int_{\mu}^{\overline{v}}vf(v)dv+\mu F(\mu)-\int_{\mu}^{\overline{v}}v\hat{f}(v)dv\\\tag{2} &=\int_{\mu}^{\overline{v}}vf(v)dv+\mu F(\mu)-\int_{\mu}^{\overline{v}}v[2f(v)F(v)]dv\\ &=\int_{\mu}^{\overline{v}}vf(v)[1-2F(v)]dv+\mu F(\mu)\geq 0\end{align} $$
where $\hat{V}$ denotes the random variable that is the maximum of two independent copies of $V$, $\hat{F}$ given by $\hat{F}(v)=[F(v)]^2$ is its distribution function and $\hat{f}$ given by $\hat{f}(v)=2f(v)F(v)$ is its density function.
Some extra details:
Let $\phi(x)=g(x)-h(x)$ where $g$ and $h$ are given by
$$g(x)=\int_{x}^{\overline{v}}vf(v)dv+\mu F(x),\qquad \text{and} \qquad h(x)=\int_{x}^{\overline{v}}v[2f(v)F(v)]dv$$
Then inequality $(1)$ is $\phi(\mu)\geq 0$. Denoting the mean of $\hat{F}$ by $\hat{\mu}$, it is easy to see that:
$$\phi(\underline{v})=\mu-\hat{\mu}<0\quad \text{and}\quad \phi(\overline{v})=\mu>0$$
Also, $\phi$ is strictly increasing:
$$\phi'(x)=(\mu-x+2xF(x))f(x)\geq xF(x)f(x)>0,\quad \forall x\in(\underline{v},\overline{v})$$
where the first inequality follows from Markov's inequality.
It follows that there exists a unique $\bar{x}\in(\underline{v},\overline{v})$ such that $$\phi(x)\lesseqqgtr0\iff x \lesseqqgtr\bar{x}.$$
Additionally $$g'(x)\gtreqqless0\iff x \lesseqqgtr\mu$$ so that the maximum of $g$ is $g(\mu)$.
Note also that integrating by parts gives
$$g(x)=\bar{v}+(\mu-x)F(x)-\int_x^{\overline{v}}F(v)dv\qquad \text{and}\qquad h(x)=\bar{v}-x[F(x)]^2-\int_x^{\overline{v}}[F(v)]^2dv$$
Some facts related to log-concavity:
- Log-concavity of $f$ implies log-concavity of $F$ and $1-F$
- If $f$ is differentiable then log-concavity of $F$ is equivalent to $f^2-Ff'\geq 0$ and log-concavity of $1-F$ is equivalent to $(1-F)f'+f^2\geq 0$
- Log-concavity of $F$ implies $F(\mu)\geq \frac{1}{e}$
- Products of log-concave functions are log concave. For example $F[1-F]$ is log-concave. -Log-concave densities are (strongly) unimodal
A proof that $\phi(\mu)\geq 0$ if $\mu\geq m$ and $\mu\geq \overline{v}/2$
Let $m$ be the median of $F$. Then
$$\phi(m)=\frac{1}{4}(2\mu-m)-\int_m^{\overline{v}}F(v)[1-F(v)]dv\geq \frac{1}{4}(2\mu-\overline{v})$$
where the last inequality follows because the integrand is at most $1/4$. Since $\phi$ is increasing and $\mu\geq m$, the result follows.
(Note that this result allows us to conclude that $\phi(\mu)\geq 0$ for all symmetric distributions.)
A lower bound for $\phi(\mu)$
From Chebyshev's integral inequality:
$$\int_{\mu}^{\overline{v}}[F(v)]^2dv\geq \frac{1}{\overline{v}-\mu}\left[\int_{\mu}^{\overline{v}}F(v)dv\right]^2$$
Thus
$$-\int_{\mu}^{\overline{v}}F(v)[1-F(v)]dv\geq -\frac{1}{\overline{v}-\mu}\int_{\mu}^{\overline{v}}F(v)dv\left(\overline{v}-\mu-\int_{\mu}^{\overline{v}}F(v)dv\right)$$
Since $\int_{\mu}^{\overline{v}}F(v)dv\leq \overline{v}-\mu$ and
$$\int_{\mu}^{\overline{v}}F(v)dv=\overline{v}-\mu F(\mu)-\int_{\mu}^{\overline{v}}vf(v)dv$$
we get
$$-\int_{\mu}^{\overline{v}}F(v)[1-F(v)]dv\geq \mu[1-F(\mu)]-\int_{\mu}^{\overline{v}}vf(v)dv.$$
It follows that $$\begin{align*} \phi(\mu)&\geq\mu F(\mu)^2+\mu[1-F(\mu)]-\int_{\mu}^{\overline{v}}vf(v)dv \\ &=\mu-\int_{\mu}^{\overline{v}}vf(v)dv-\mu F(\mu)[1-F(\mu)]\\ \tag{3} &=\int_{\underline{v}}^{\mu}vf(v)dv-\mu F(\mu)[1-F(\mu)]\\ &=\mu F(\mu)^2-\int_{\underline{v}}^{\mu}F(v)dv \end{align*}$$
The lower bound above seems to be nonnegative for the beta distributions and Kumaraswamy distributions. It is exactly zero for the uniform distribution on $[0,1]$.
A proof that $\phi(\mu)\geq 0$ when $f$ is increasing on $[\underline{v},\mu]$ and $F(\mu)\geq 1/2$
Since $F(\mu)[1-F(\mu)]\leq 1/4$, the lower bound $(3)$ gives the following sufficient condition for $\phi(\mu)\geq 0$: $$\int_{\underline{v}}^{\mu}vf(v)dv\geq\frac{1}{4}\mu$$
If $f$ is increasing on $[\underline{v},\mu]$ and $F(\mu)\geq 1/2$ then,
$$\int_{\underline{v}}^{\mu}vf(v)dv\geq \frac{\mu+\underline{v}}{2}F(\mu)\geq \frac{1}{4}\mu$$
where the first inequality comes from by Chebyshev's integral inequality.
